When P is high, its voltage is 4.5 V.
M must go below 2.5 V to make P go low at 0.5 V.
Use calculations to show that L has to be taken below 1.1 V to make P go low.

My calculations:

4.5V (P) – 1.1V (L) = 3.4V

3.4V/(330k + 470k) = 4.25uA

4.25uA*330k = 1.4V

1.1V (L) + 1.4V = 2.5V = M

Right?

 

You cannot use 1.1V in your calculation because that is the voltage you are supposed to calculate.

Because of the extremely high input impedance of the CMOS gate, the input current is basically 0. Therefore your problem can be simplified to the finding of an input voltage that gives point 'M' 2.5V.

Try again.

 

(4.5V - 2.5V)/470k = 4.25uA

2.5V/330k = 7.57uA

7.57uA - 4.25uA = 3.32uA

3.32uA*330k = 1.1V

 

(4.5V - 2.5V)/470k = 4.25uA

Correct.

This is the only current in the loop. There is no other current path.

2.5V/330k = 7.57uA

Wrong. The current in 330K is the same as 470K because they are in series. The current in 470K has already shown to be 4.25uA, so current in 330K is also 4.25uA.

Voltage drop across 330K is thus 4.25uA * 330K = 1.403V

With M = 2.5V, it follows that the input voltage must be 2.5 - 1.403 = 1.098V.

 

Thank you.